Electrical Sba (cape) Transmission Power Line Test

Electrical & Electronics Technology S.B.A. (CAPE) Model Power Line

Name:

Tarick Edwards

Date:

06 Feb 2009

Lecturer: Mr. A. Clarke Reg#:

1000101042

Territory: Jamaica Table of Content Acknowledgement ………………………………………………………..2 Purpose of the report…………………………………………………….3

Aim……………………………………………………………………….……….4 Scope& Specification……………………………………………………..5 Theory……………………………………………………………………………6 Methodology……………………………………………………………......8 Design & Construction detail...…………………………………..….9 Summary of Tests and Troubleshooting procedure…….…18 Tests and Troubleshooting Results………………………………..19 Discussion……………………………………………………………………..20 Constraints & Difficulties/limitations………………………....…21 Conclusion/Recommendations…………………………………..…22

Acknowledgement I sincerely Mr. Clarke guide us in construction of the project, and also in the assembling of the information gathered in the correct format. I also wish to give thanks to the pass student for their marvellous pass project. 1

Purpose of the report The purpose of this project is to satisfy the requirements of CAPE Electrical and Electronic Technology.

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Aim The aim of this project is to construct a model AC power line at high voltage in order to investigate the power losses in the transmission lines. The AC model will deliver up to 12 volts at the rated output. Also a DC power line to be compared with the AC lines. This project also gives us a lot of practical experience working with the device we leant by theory.

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Scope and Specifications This project is based on the creation of a Power Transmission System. This Power Transmission should entail: Two Transformers, one bulb, Length Constantan wire Transformers Step down Transformer – 110V/12V 50Hz, KVA Step up Transformer – 12V/240 50Hz, 4KVA Step down Transformer – 240V/12V 50Hz, 4KVA Wire ➢ Constantan Wires length – 1m each ➢ Daimeter-0.32mm ➢ Resistivity- 49x10-8 4

➢ Wire gauge- 30 SWG

Theory Power lines are overhead wires supported by high towers that transmit electric energy from power supplies or plants. The centre strands of power lines are made of steel to give them strength and the outer strands are made of aluminum because of its lightness and ability to carry current. The wires are insulated from the towers by porcelain insulators to prevent the loss of electric energy. Electric power transmission is a process in the delivery of electricity to consumers and also the bulk transfer of electric power. D.C. power transmission is the distribution of power using direct current. Direct current can be transmitted at high voltages but to lower it for consumers would be costly. A.C. power transmission is the distribution of power using alternating current. A.C. voltages has the advantage of increasing and decreasing in value more readily than that of the (D.C).Transmission efficiency is improved by increasing the voltage using a step up transformer which reduces the current in the conductors while keeping the power transmitted nearly equal to the power input. The reduced current flowing through the conductor reduces the losses in the conductor and since, according to Joule's Law, the losses are 5

proportional to the square of the current, halving the current makes the transmission loss one quarter the original value. In alternating current systems, energy loss across power lines is reduced because transformers make it possible to raise the A.C. voltage to very high values. These high voltages allow the same level of electric power made available at a lower current. This results in less power loss, smaller transmission cables and higher efficiency. In addition to stepping up or raise the voltage for long distance transmission, transformers also step-down or lower the voltage to the requirements of the load. PLoss = I2R The transformer is based on two principles: firstly that an electric current can produce a magnetic field (electromagnetism) and secondly that a changing magnetic field within a coil of wire induces a voltage across the ends of the coil (electromagnetic induction). By changing the current in the primary coil, it changes the strength of its magnetic field; since the changing magnetic field extends into the secondary coil, a voltage is induced across the secondary.

Diagram showing a simplified transformer design

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A current passing through the primary coil creates a magnetic field. The primary and secondary coils are wrapped around a core of very high magnetic permeability, such as iron; this ensures that most of the magnetic field lines produced by the primary current are within the iron and pass through the secondary coil as well as the primary coil. The voltage induced across the secondary coil may be calculated from Faraday's law of induction, which states that: where VS is the instantaneous voltage, NS is the number of turns in the secondary coil and Φ equals the magnetic flux through one turn of the coil. If the turns of the coil are oriented perpendicular to the magnetic field lines, the flux is the product of the magnetic field strength B and the area A through which it cuts. The area is constant, being equal to the cross-sectional area of the transformer core, whereas the magnetic field varies with time according to the excitation of the primary.

Methodology and Approach 1) The information on Power Transmission Systems was gathered 2) The value of various circuit components was calculated (e.g. resistance of wire). 3) The equipment and components needed were then gathered and assembled. 4) The current and voltage readings of both circuits were measured (for the 2V D.C. and 12V A.C.) 5) The current and voltage readings of the lamps and the voltage across the transmission line were recorded. 6) The power losses were then calculated.

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Design and Construction details These Materials were gathered: 12-240 step up Transformer, 240-12 step down Transformer, 110-12 step down transformer, switch (SPDT), 12 lamp, 2 Piece 1m constantan wire, multi-meter, 12 DC supply

12V D.C. The lamp was connected across the 12V D.C. Supply as shown in fig.1

8

R1 4.87

Source 12 V 2

Lamp 12V, 50W

R4

1

4.99

12V A.C. 1) The lamp was connected to the secondary winding of the (110V-12V)

transformer. 2) The primary winding of the transformer is connected to a 12V A.C. Supply. 3) (step down transformer, 110V/12V) as shown in fig.2 1

3

Transformer Source

120 V 50 Hz 0Deg NLT_PQ_4_120 5

R1

2

4.87

6

R2

Lamp 12V, 50W 4

4.99 240V A.C. 1) The 110V/12V transformer was connected to one of the bulbs 2) The bulb was then connected across a 12V/240V transformer 3) The transmission line was then connected to the transformer 4) The other ends of the transmission line was then connected to a 240V/12V transformer 5) The transformer was then connected to a 12V AC power supply. 9

(Shown in fig.3)

1

7

Transformer1 Source

12 V 50 Hz 0Deg 4

NLT_PQ_4_12

R1

5

Transformer2

4.87

6

R2

2

Lamp 12V, 50W 8

NLT_PQ_4_12

3

4.99

Calculations 12 Volt AC Supply

Expected Values

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Expected Resistance of Transmission lines

Side Notes / Key

R= ρl/A R = 49 x 10-8 x 2m

ρ –Resistivity of constantan- 4.9x10-7 Ωm

8.042 x 10-8m2

Diameter of wire used - 0.32 x 10-3 m

R = 12.19Ω

Radius of wire – 0.16 x 10-3m Expected Bulb Resistance – 2.88 Ω

The Area of constantan wire

Expected Vdrop in transmission Lines(9.75v)

A = Π r2

Expected

A = Π x (0.16 x 10-3)2

V

supply

-12v

Expected wire resistance (R

A = Π x 2.56 x 10-8

Line)

- 12.19Ω

Actual Bulb Resistance - 0.6Ω

A = 8.042 x 10-8m2

Actual Current Supplied - 1.03A Actual Resistance of Lines - 9Ω

Expected Current

Actual voltage supplied - 11.28v

I = V/ RT

Actual voltage across lamp - 2v

I = 12 / 15.07

Power delivered = Output Power -2.06W

I = 0.80A

Power Supplied = Input Power- 11.62W

RT = 12.19 + 2.88 = 15.07 Ω Expected voltage Drop in Lines V=IxR

line

V = 0.8A x 12.19 Ω V = 9.75v )2 / R

P

Loss

Expected =( V

P

Loss

= (9.75)2 / 12.19

P

Loss

= 7.80W

drop

Line

Expected Voltage at Load 11

V

Lamp

=V

-V

V

Lamp

= 12 – 9.75

V

Lamp

= 2.25v

supply

drop in Lines

Measured Values Power loss measured in Transmission Lines P

Loss

measured = I2R

P

Loss

measured = 1.032 x 9Ω

P

Loss

measured = 9.55W

Power Measured at Lamp/Load. P

Delivered to Lamp

=I

P

Delivered to Lamp

P

Delivered to Lamp

Supplied

xV

Delivered to lamp

= 1.03A x 2V = 2.06W

Measured Input Power P

Generated

= I

P

Generated

= 1.03A x 11.28V

P

Generated

Supplied

supplied

= 11.62W

%η = P Delivered P

xV

x 100

Generated

%η = 2.06

x 100

11.62

%η = 17.7 12

Where

η is the Efficiency. 12 Volt Dc Supply Side Notes for calculations

Total Resistance - 15.07Ω Voltage Supplied - 12v Current in circuit 1.12A Expected Values

Actual voltage across lamp 2.23v

Expected Current

Expected wire resistance Power

I = V/ RT

-2.24W

I = 12 / 15.07 = 0.80A RT = 12.19 + 2.88 = 15.07 Ω Expected voltage Drop in Transmission Lines V=IxR

Line

V = 0.8A x 12.19 Ω V = 9.75v

)2 / R

P

Loss

Expected =( V

P

Loss

= (9.75)2 / 12.19

P

Loss

= 7.80W

drop

Expected Voltage at Load V

Lamp

=V

V

Lamp

= 12 – 9.75

V

Lamp

= 2.25v

supply

-V

drop in Lines

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Measured Values

Power measured in Transmission lines

P

Loss

measured = I2R

P

Loss

measured = 1.122 x 9Ω

P

Loss

measured = 11.28W

Power Measured at the Load P

Delivered to Lamp

=I

P

Delivered

= 1.12A x 2V

P

Delivered

= 2.24W

Supplied

xV

Delivered to lamp

Input Power P

Generated

= I

P

Generated

= 1.12A x 12V

P

Generated

Supplied

supplied

= 13.44W

%η = P Delivered P

xV

x 100

Generated

%η = 2.24

x 100

13.44

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%η = 18.15

High Voltage Transmission (110V AC)

Side Notes for Calculations Expected Values V p – Primary voltage in Step up Transformer 9V Expected power loss

P

loss

= (Vdrop) 2 /R

P

loss

= 2.072/12.19

line

= 0.352w

I p – Primary Current in Step up Transformer 2A Vs – Voltage in secondary of Step up Transformer 110V Is – Current in Secondary of Step up Transformer = 0.164A

Expected current in transmission line : V

p

/ Vs = Is /I

p

9/110 = Is/2 Is = 0.082 x 2 Is = 0.164A * Is is also the voltage transmitted to the Transmission lines Expected voltage drop Vdrop = I x R

line

= 0.164 x 12.19 = 2V – Expected = 107.93V 15

Expected V

= Expected V

After T line

bef. T line

– Vdrop in line =110- 2.07 = 107.93V

Measured Values

Measured V

bef. T line

= 45.8V

Measured VAfter T line = 43.8

Measured Voltage drop in Transmission lines V

drop

=V

V

drop

= 45.8 – 43.1

V

drop

= 2.7v

before T lines

–V

after T Lines

Measured current in lines – 0.17A

Power Loss Measured Ploss = I2R Ploss = 0.172 x 9Ω Ploss = 0.260W

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Testing & Troubleshooting Procedures

1) The Multimeter was used to test the Resistance of the Length

of Constantan. 2) It was then used to test the current and voltage through/across each bulb for the D.C. Supply. 3) The digital multi-meter was also used to measure the voltage and current across/through the bulb when connected as shown in the schematics for the A.C. Power Transmissions. 4) The digital multi-meter was again used to measure the voltage across the power line in fig.3.

Testing & Troubleshooting Results 17

12 Volt AC supply

Parameters

Value Expected

Value Measured

Error

0.80A

1.03A

0.23

12V

11.28v

0.72

Voltage Drop

9.75V

9.76V

0.01

Power Loss

7.80W

9.55W

1.75

Resistance of Lines

12.19Ω

9Ω

3.19

Current Supply Voltage

12 Volt Dc Supply Parameters

Value Expected

Value Measured

Error

Current

0.8A

1.12A

0.32

Supply Voltage

12V

12V

-

Voltage Drop

9.75V

9.55V

0.20

Power Loss

7.80W

11.28W

3.48

12.19 Ω

9Ω

3.19

Resistance of Lines

240v ac Supply with Mounted Transformer at Both Ends Parameters

Value Measure d

Value Expected

Error

Primary Voltage

9.27v

9V

0.27

Secondary Voltage

45.8v

110V

64.2

Primary Current

2.14A

2.0A

0.14

Secondary Current

0.17A

0.164A

0.006

Power Loss in lines

0.26W

0.35W

0.09

Voltage Delivered to user

1.85V

1.53V

0.32

Transformer 1

18

Current Delivered to user

1.58A

2A

0.42

Discussion The circuits were constructed using the required components; hence, our requirements were achieved. The resistances of both lengths of constantan wires were both taken and recorded. The resistance of the lamp was also measured and recorded. The 12 AC circuit was constructed using a step down transformer (120V-12V). The primary windings of the transformer are connected to the power supply, and the secondary windings connected to two 1 meter length of constantan wire. The constantan wires are then connected to the lamp. Ideally, the reading on the secondary windings would be 12v, but it was measured to be 11.28V. The voltage drops across the constantan wires were measured. The current in the circuit was 1.03A; this is constant since it is a series circuit. The lamp had small illumination due to the fact that most of the voltage drops across the constantan wires. But it would definitely show better in a dark area. After the circuit was powered up for a while, the temperature of the constantan wires began to increase. After completing the test of the 12V AC Circuit, the circuit was disassembled and reconstructed into a high voltage transmission line. With the use of two transformers refer to diagram below. 1

7

Transformer1 Source

12 V 50 Hz 0Deg 4

NLT_PQ_4_12

R1

5

Transformer2

4.87

6

R2

2

Lamp 12V, 50W 8

NLT_PQ_4_12

3

4.99

The voltage across the secondary winding of the transformer was measured. A voltage 43.1V was measured at primary end of transformer2. The voltage at the load was found to be1.81V. The voltage drop across the transmission line R1 and R2 were both measured; both result in a value of 0.6V. The current was then measured and a value of 0.17A was obtained. The circuit has the least power lost of all. The lamp light could be seen clearly. That is why Power Company transmits electricity at high Voltage and low

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current. It is also less expensive to transmit at low current (smaller wires can be used). After completing the circuit illustrated of the circuit above. A 12V DC circuit was assembled. The 12V DC supply was connected across the 12V lamp. The voltage across the transmission line was measured and recorded to be 9.55V, it was assumed to be 9.75V. The circuit was open and a multi-meter used to test the current in the circuit, which was found to be 1.12A. This circuit has the highest power loss. Hence, the lamp did not light. That is why this is a poor method of transmitting electricity. It is expensive and inefficient to distribute due to high temperature, overheat in the wire. As experienced in the experiment. It is also difficult to convert from Dc to Ac.

Constraints & Difficulties/ Limitations 1) It was difficult to source the constantan wire. 2) The temperature of the room was constantly changing thus affecting the results. (Drawn from ohms law). 3) It was difficult to get a 12 volt AC supply. (An 110V to 12V step-down transformer was used). 4) The AC voltage fluctuated thus causing a variety of voltages and currents readings on the multi-meter. 5) The someof voltages, currents, resistance, and reading on the multi-meter never kept still and hence were estimated.

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Conclusion

The Power loss in a Power Transmission line is I2R, where I is the current through the power line and R is the resistance of the line. It can also be concluded that the larger the voltage being transmitted, the smaller the power loss. That is why power companies transmit electricity at high voltages. AC and also recommended over DC, because, AC transmission is less expensive transmit as I can be stepped up and down easily by a transformer. AC is also preferable because it is easily converted.

Recommendations 1) It is recommended that A.C. is used over D.C. for transmission as A.C. is less expensive to transmit as it can be stepped up by the use of a transformer. 2) It is also recommended that when transmitting electrical power, it should be transmitted at high voltages so as to limit the power loss.

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